Why $\mathbb{Z}[\theta]\,/\,\mathcal{P}$ is an algebraic extension over $\mathbb{Z}/p\mathbb{Z}$? 
If $f(x)$ is a monic, irreducible polynomial in $\mathbb{Z}[x]$ with $\theta\in\mathbb{C}$ as root, why $\mathbb{Z}[\theta]\,/\,\mathcal{P}$ is an algebraic extension over $\mathbb{Z}/p\mathbb{Z}$?

I'm interested in a simple explanantion, if there exists, without using the Krull dimension. 
I already proved that $\mathbb{Z}/p\mathbb{Z} \subset \mathbb{Z}[\theta]\,/\,\mathcal{P}$.
 A: Note that if $R \subset S$ is an inclusion of integral domains, then for any prime ideal $\mathcal{P} \in S$, $\mathcal{P} \cap R$ is a prime ideal of $R$. 
Here $R =  \mathbb{Z}, S = \mathbb{Z}[\theta]$ and $\mathcal{P} \cap \mathbb{Z} = p \mathbb{Z}$ for some prime number $p$, since it is a prime ideal of $\mathbb{Z}$. 
Thus $\mathbb{Z}/p\mathbb{Z} \subset \mathbb{Z}[\theta]\,/\,\mathcal{P}$, which is almost a proof of what you want. 
All you need is saying that since $\theta$ is an algebraic integer, $\mathbb{Z}[\theta]$ is a finitely generated free $\mathbb{Z}$-module (its elements are of the form $\sum_{n=0}^{deg(f)-1} a_n \theta^n, a_n \in \mathbb{Z}$), 
so that $$\mathbb{Z}[\theta]\,/\,p \mathbb{Z}[\theta]= \{ \sum_{n=0}^{deg(f)-1} a_n \theta^n, a_n \in \mathbb{Z}/p \mathbb{Z} \}$$ is a finite ring. Clearly it contains $\mathbb{Z}[\theta]\,/\,\mathcal{P}$,
therefore $k = \mathbb{Z}[\theta]\,/\,\mathcal{P}$ is a finite integral domain, thus a finite field. 
Since $k$ contains $\mathbb{F}_p$, you get that $k/\mathbb{F}_p$ is a finite (thus algebraic) field extension.
Indeed $k = \mathbb{F}_p[\overline{\theta}]$ where $\overline{\theta}$ is one of the root of $f \bmod p$ (and there is one prime ideal above $p$ for each of those different root)
